proof of Hahn-Banach theorem
Consider the family of all possible extensions of , i.e. the set of all pairings where is a vector subspace of containing and is a linear map such that for all and for all . is naturally endowed with an partial order
relation
: given we saythat iff is an extension of that is and for all .We want to apply Zorn’s Lemma to so we are going to prove that every chain in has an upper bound.
Let be the elements of a chain in . Define . Clearly is a vector subspace of and contains . Define by “merging” all ’s as follows. Given there exists such that : define . This is a good definition since if both and contain then in fact either or .Notice that the map is linear, in fact given any two vectors there exists such that and hence .The so constructed pair is hence an upper bound for the chain because is an extension of every .
Zorn’s Lemma then assures that there exists a maximal element . To complete
the proof we will only need to prove that .
Suppose by contradiction that there exists . Then consider the vector space
( is the vector space generated by and ).Choose
We notice that given any it holds
i.e.
in particular we find that and for all it holds
Define as follows:
Clearly is a linear functional.We have
and by letting by the previous estimates on we obtain
and
which together give
and hence
So we have proved that and which is a contradiction.